Answer:
Final kinetic energy, [tex]K_f=4000\ MJ[/tex]
Explanation:
It is given that,
Initial gravitational potential energy of the satellite, [tex]U_i=5200\ MJ[/tex]
Initial kinetic energy of the satellite, [tex]K_i=4800\ MJ[/tex]
Final gravitational potential energy of the satellite, [tex]U_f=6000\ MJ[/tex]
Let [tex]K_f[/tex] is the final kinetic energy of the satellite at that point. Using the conservation of energy to find it as the energy of the satellite remains constant.
[tex]U_i+K_i=U_f+K_f[/tex]
[tex]K_f=U_i+K_i-U_f[/tex]
[tex]K_f=5200+4800-6000[/tex]
[tex]K_f=4000\ MJ[/tex]
So, the final kinetic energy of the satellite at that point is 4000 MJ. Hence, this is the required solution.